How many grams of NH4Cl must be added to 0.100 L of 0.441 M NH3 to produce a buffer solution with pH= 9.38 ? pKa (NH4Cl) = 9.24?

NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) …. pKa = 9.24

pH = pKa + log([NH₃]/[NH₄⁺])
pH = pKa + log([NH₃]/[NH₄Cl])
9.38 = 9.24 + log(0.441/[NH₄Cl])
log(0.441/[NH₄Cl]) = 0.14
log([NH₄Cl]/0.441) = -0.14
[NH₄Cl]/0.441 = 10⁻⁰˙¹⁴
[NH₄Cl] = 0.441 × 10⁻⁰˙¹⁴ mol/L = 0.319 mol/L

No. of moles of NH₄Cl = (0.319 mol/L) × (0.100 L) = 0.0319 mol

Molar mass of NH₄Cl = (14.0 + 1.0×4 + 35.5) g/mol = 53.5 g/mol
Mass of NH₄Cl must be added = (0.0319 mol) × (53.5 g/mol) = 1.71 g